exercice Jupyter

module2/exo1/toy_notebook_fr.ipynb

 {
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+ "cells": [
+  {
+   "cell_type": "markdown",
    "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "# 1 À propos du calcul de *$\\pi$* "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "## 1.1 En demandant à la lib maths "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "Mon ordinateur m’indique que π vaut *approximativement*"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "outputs": [],
+   "source": [
+    "from math import *\n",
+    "print(pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "## 1.2 En utilisant la méthode des aiguilles de Buffon"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "Mais calculé avec la **méthode** des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme **approximation** :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "np.random.seed(seed=42)\n",
+    "N = 10000\n",
+    "x = np.random.uniform(size=N, low=0, high=1)\n",
+    "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
+    "2/(sum((x+np.sin(theta))>1)/N)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
+   "source": [
+    "## 1.3 Avec un argument \"fréquentiel\" de surface"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction\n",
+    "sinus se base sur le fait que si X ∼ U(0,1) et Y ∼ U(0,1) alors P[X^2 + Y^2 ≤ 1] = π/4 (voir\n",
+    "[méthode de Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)). Le code suivant illustre ce fait :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "%matplotlib inline\n",
+    "import matplotlib.pyplot as plt\n",
+    "\n",
+    "np.random.seed(seed=42)\n",
+    "N = 1000\n",
+    "x = np.random.uniform(size=N, low=0, high=1)\n",
+    "y = np.random.uniform(size=N, low=0, high=1)\n",
+    "\n",
+    "accept = (x*x+y*y) <= 1\n",
+    "reject = np.logical_not(accept)\n",
+    "\n",
+    "fig, ax = plt.subplots(1)\n",
+    "ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n",
+    "ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
+    "ax.set_aspect('equal')"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "    Il est alors aisé d’obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois,en moyenne, X^2+Y^2 est inférieur à 1 :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "4*np.mean(accept)"
+   ]
+  }
+ ],
+ "metadata": {
+  "hide_code_all_hidden": false,
   "kernelspec": {
    "display_name": "Python 3",
    "language": "python",
@@ -16,10 +156,9 @@
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-